Area

SAT Subject Test in Math I · Learn by Concept

Help Questions

SAT Subject Test in Math I › Area

1 - 10

Find the area of a kite with diagonal lengths of and .

CORRECT

Explanation

Write the formula for the area of a kite.

$A= \frac{d1\cdot d2}{2}$

Plug in the given diagonals.

$A= \frac{(a+b)\cdot (2a-2b)}{2}$

Pull out a common factor of two in and simplify.

$A= \frac{(a+b)\cdot2(a-b)}{2}$

Use the FOIL method to simplify.

Triangle

Note: Figure NOT drawn to scale.

Refer to the figure above, which shows a square inscribed inside a large triangle. What percent of the entire triangle has been shaded blue?

$44\frac{4}{9} %$

CORRECT

$33 \frac{1}{3} %$

Insufficient information is given to answer the question.

Explanation

The shaded portion of the entire triangle is similar to the entire large triangle by the Angle-Angle postulate, so sides are in proportion. The short leg of the blue triangle has length 20; that of the large triangle, 30. Therefore, the similarity ratio is . The ratio of the areas is the square of this, or $3^{2}:2^{2}$ , or .

The blue triangle is therefore $\frac{4}{9}$ of the entire triangle, or $\frac{4}{9} \times 100 = 44 \frac{4}{9} %$ of it.

Find the area of a circle with a diameter of .

$64\pi$

CORRECT

$32\pi$

$16\pi$

$81\pi$

$256\pi$

Explanation

Write the formula for the area of a circle.

$A_{circle}= \pi r^2 = \frac{\pi d^2}{4}$

Substitute the diameter and solve.

$\frac{\pi (16)^2}{4} = \frac{\pi (16)(16)}{4}=\pi (16)(4)= 64\pi$

On the XY plane, line segment AB has endpoints (0, a) and (b, 0). If a square is drawn with segment AB as a side, in terms of a and b what is the area of the square?

CORRECT

$A=\frac{1}{2} (a + b)$

Cannot be determined

Explanation

Since the question is asking for area of the square with side length AB, recall the formula for the area of a square.

$A=\text{side}^2$

Now, use the distance formula to calculate the length of AB.

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

let

$\(x_1,y_1)=(0,a) \(x_2,y_2)=(b,0)$

Now substitute the values into the distance formula.

$\d=\sqrt{(b-0)^2+(0-2)^2} \d=\sqrt{b^2+a^2}$

From here substitute the side length value into the area formula.

$\A=\text{side}^2 \A=\left(\sqrt{b^2+a^2}\right)^2 \A=b^2+a^2$

Right_triangle

Note: Figure NOT drawn to scale.

Refer to the above diagram. Give the ratio of the area of $\Delta BDC$ to that of $\Delta ADB$ .

CORRECT

Insufficient information is given to answer the question.

Explanation

, as the length of the altitude corresponding to the hypotenuse, is the geometric mean of the lengths of the parts of the hypotenuse it forms; that is, it is the square root of the product of the two:

$BD =\sqrt{ \left ( AD\right ) \left ( DC\right )} = \sqrt{6 \cdot 24} = \sqrt{144} = 12$ .

The areas of $\Delta BDC$ and $\Delta ADB$ , each being right, are half the products of their legs, so:

The area of $\Delta ADB$ is $\frac{1}{2} \times (AD) (BD) = \frac{1}{2} \times 6 \times (BD)= 3(BD)$

The area of $\Delta BDC$ is $\frac{1}{2} \times (CD) (BD) = \frac{1}{2} \times 24 \times (BD)= 12(BD)$

The ratio of the areas is $\frac{12 (BD)}{3 (BD)} = \frac{12 }{3 } =\frac{4}{1}$ - that is, 4 to 1.

Give the area of $\bigtriangleup ABC$ to the nearest whole square unit, where:

CORRECT

Cannot be determined

Explanation

The area of a triangle, given its three sidelengths, can be calculated using Heron's formula:

$A = \sqrt{s(s-a)(s-b)(s-c)}$ ,

where , , and are the lengths of the sides, and $s = \frac{a+b+ c}{2}$ .

Setting, , , and ,

$s = \frac{13+17+25}{2} = \frac{55}{2} = 27.5$

and, substituting in Heron's formula:

$A = \sqrt{27.5(27.5-13)(27.5-17)(27.5-25 )}$

$= \sqrt{27.5(14.5)(10.5)(2.5 )}$

$= \sqrt{10,467.1875}$

$\approx 102$

Give the area of $\bigtriangleup ABC$ to the nearest whole square unit, where:

$m \angle A = 57^{\circ }$

CORRECT

Explanation

The area of a triangle with two sides of lengths and and included angle of measure $\gamma$ can be calculated using the formula

$A = \frac{1}{2} ab \sin \gamma$ .

Setting , , and $\gamma = m \angle A = 57^{\circ }$ , then evaluating :

$A = \frac{1}{2} (19 )(25)\sin 57^{\circ }$

$A = \frac{1}{2} (19 )(25) (0.8387)$

$A \approx 199$ .

Find the area of a kite if the diagonal dimensions are and .

$\frac{x^2-3x-18}{2}$

CORRECT

$(x^2-3x-18)\sqrt2$

$\frac{x^2+3x-18}{2}$

$\frac{x^2-3x+18}{2}$

Explanation

The area of the kite is given below. The FOIL method will need to be used to simplify the binomial.

$A=\frac{d1\cdot d2}{2}=\frac{(x+3)(x-6)}{2}= \frac{x^2-6x+3x-18}{2}$

$A= \frac{x^2-3x-18}{2}$

The diagonals of a kite are $\sqrt10:cm$ and $\sqrt5:cm$ . Find the area.

$\frac{5\sqrt2}{2}:cm^2$

CORRECT

$\frac{2\sqrt5}{2}:cm^2$

$5\sqrt2:cm^2$

$\sqrt5:cm^2$

Explanation

The formula for the area for a kite is

$A=\frac{d_1\cdot d_2}{2}$ , where and are the lengths of the kite's two diagonals. We are given the length of these diagonals in the problem, so we can substitute them into the formula and solve for the area: